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Question Number 139918 by mathsuji last updated on 02/May/21

1+cos2°+cos4°+cos6°+...+cos88°

1+cos2°+cos4°+cos6°+...+cos88°

Answered by mr W last updated on 02/May/21

Σ_(k=0) ^(44) (cos ((2kπ)/(180))+i sin ((2kπ)/(180)))=Σ_(k=0) ^(44) e^(((2kπ)/(180))i)   =((1−e^(((2πi)/(180))×45) )/(1−e^((2πi)/(180)) ))=((1−i)/((1−cos (π/(90)))−isin (π/(90))))  =(((1−i)[(1−cos (π/(90)))+i sin (π/(90))])/((1−cos (π/(90)))^2 +sin^2  (π/(90))))  =((1−cos (π/(90))+ sin (π/(90))−i(1−cos (π/(90))+sin (π/(90))))/((1−cos (π/(90)))^2 +sin^2  (π/(90))))  =((1−cos (π/(90))+ sin (π/(90))−i(1−cos (π/(90))+sin (π/(90))))/(2(1−cos (π/(90)))))    1+cos 2°+cos 4°+...+cos 88°=  =Σ_(k=0) ^(44) cos ((2kπ)/(180))  =((1−cos (π/(90))+ sin (π/(90)))/(2(1−cos (π/(90)))))  =(1/2)(1+cot (π/(180)))  =(1/2)(1+cot 1°)

44k=0(cos2kπ180+isin2kπ180)=44k=0e2kπ180i=1e2πi180×451e2πi180=1i(1cosπ90)isinπ90=(1i)[(1cosπ90)+isinπ90](1cosπ90)2+sin2π90=1cosπ90+sinπ90i(1cosπ90+sinπ90)(1cosπ90)2+sin2π90=1cosπ90+sinπ90i(1cosπ90+sinπ90)2(1cosπ90)1+cos2°+cos4°+...+cos88°==44k=0cos2kπ180=1cosπ90+sinπ902(1cosπ90)=12(1+cotπ180)=12(1+cot1°)

Commented by mathsuji last updated on 02/May/21

thank you very much Sir..

thankyouverymuchSir..

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